Question

The reaction A + 2B → C proceeds according to the rate law, rate = k[A][B] with k = 6.05 x 10-3 M-1 min-1. Assume that at the start of a particular reaction, [A] = 0.275 M, [B] = 0.0950 M. What is the initial rate of reaction? What is the rate when half of B has reacted? Give units! [Remember: ln(ab) = ln(a) + ln(b)]

Answer 1

SOLUTION:

Given reaction:

A+ 2B → C

Rate law for this reaction:

rate = kA.B]

Initially:

k = 6.05 x 10^-3 M^-1 min^-1

[A]_o = 0.275 M

[B]_o = 0.0950 M

Initial rate is given by:

rate = k[A]o.[B]. (6.05x10-3 M-' min-1)x (0.275 M)x (0.0950 M) %3D

rate = 1.581 x 10-4 M/min +(answer)

When half of B is reacted:

[B] = [B]_o- [B]_o/2 = (0.0950 M) - (0.0950 M)/2 = 0.0475 M

From the given reaction, according to stoichiometry (0.0475 M)/2 of A will be consumed = 0.02375 M

[A] = 0.275 M - 0.02375 M = 0.25125 M

Therefore, rate is given by:

rate = k[A].[B] = (6.05x10-3 M-1 min-)x(0.25125 M)x (0.0475 M) %3D

​​​​​​​

rate = 7.22 x 10-5 M/min (answer)